Counting involutions on multicomplex spaces

Nicolas Doyon Nicolas Doyon and William Verreault: Département de mathématiques et de statistique, Université Laval, Quebec City, QC, G1V 0A6, Canada Nicolas.Doyon@mat.ulaval.ca, william.verreault.2@ulaval.ca , Pierre-Olivier Parisé Pierre-Olivier Parisé: Department of Mathematics, University of Hawai’i at Manoa, Honolulu, Hawai’i, United States, 96822 parisepo@math.hawaii.edu and William Verreault

Abstract.

We show that there is a bijection between real-linear automorphisms of the multicomplex numbers of order $n$ and signed permutations of length $2^{n-1}$ . This allows us to deduce a number of results on the multicomplex numbers, including a formula for the number of involutions on multicomplex spaces which generalizes a recent result on the bicomplex numbers and contrasts drastically with the quaternion case. We also generalize this formula to $r$ -involutions and obtain a formula for the number of involutions preserving elementary imaginary units. The proofs rely on new elementary results pertaining to multicomplex numbers that are surprisingly unknown in the literature, including a count and a representation theorem for numbers squaring to $\pm 1$ .

Key words and phrases:

Clifford algebras, real algebras, multicomplex numbers, bicomplex numbers, signed permutations, involutions.

2020 Mathematics Subject Classification:

Primary: 05Axx; Secondary: 05A05, 15A66, 13J30.

1. Introduction

Let $f$ be a function on a ring $A$ containing $\mathbb{R}$ as a subfield. We say that $f$ is an involution of $A$ if $f$ is a real-linear ring homomorphism satisfying $f(f(a))=a$ for any $a\in A$ .

When $A$ is the complex plane, it is easy to show that the only involutions $f:\mathbb{C}\rightarrow\mathbb{C}$ are $f(z)=z$ and $f(z)=\overline{z}$ , the usual complex conjugate. However, for other rings, the situation might change drastically. For example, if $A$ is the field of quaternions, then we know from the work in [12, 14] that there are infinitely many involutions. If $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ is a quaternion with the usual rules

\displaystyle\mathbf{i}\mathbf{j}=-\mathbf{j}\mathbf{i}=\mathbf{k},\quad\mathbf{j}\mathbf{k}=-\mathbf{k}\mathbf{j}=\mathbf{i},\quad\mathbf{k}\mathbf{i}=-\mathbf{i}\mathbf{k}=\mathbf{j}

and $\mathbf{i^{2}}=\mathbf{j^{2}}=\mathbf{k^{2}}=-1$ , then any involution is given by $f_{\mu}(q)=\mu q\mu$ where $\mu=a_{0}\mathbf{i}+b_{0}\mathbf{j}+c_{0}\mathbf{k}$ with $a_{0}^{2}+b_{0}^{2}+c_{0}^{2}=1$ . This is a drastic change compared to the situation of the complex numbers where there are only two involutions of the complex plane.

In a recent work [20], the second author replaced quaternions by the commutative ring of bicomplex numbers. The bicomplex numbers are defined similarly to the quaternions. A bicomplex number $s$ is defined as $s=a+b\mathbf{i_{1}}+c\mathbf{i_{2}}+d\mathbf{i_{1}}\mathbf{i_{2}}$ with the rules

\displaystyle\mathbf{i_{1}}\mathbf{i_{2}}=\mathbf{i_{2}}\mathbf{i_{1}},\quad\mathbf{i_{1}^{2}}=\mathbf{i_{2}^{2}}=-1.

Bicomplex numbers are usually denoted by $\mathbb{M}\mathbb{C}(2)$ or $\mathbb{B}\mathbb{C}$ . Main references for these are [17, 21]. If we ask how many involutions there are of the bicomplex numbers, then the situation is much more similar to the complex plane case, since there are only 6 involutions of the bicomplex numbers. Surprisingly, only 4 of these involutions had been found in the literature before. For more details on this, see [20, Theorem 1].

The goal of this paper is to extend the result from [20] to the multicomplex numbers of order $n\geq 1$ , denoted by $\mathbb{M}\mathbb{C}(n)$ . The multicomplex numbers are a generalization of the complex numbers and the bicomplex numbers to higher dimensions. More and more efforts are put in the development of the theory of multicomplex numbers in the past decades. For example, they are used to introduce generalizations of concepts from real and complex analysis, e.g., multicomplex fractional operators [3], multicomplex hyperanalytic functions [26], Laurent series [16], Riemannian and semi-Riemannian geometry [27], and multicomplex holomorphic functions [24]. They were also used to generalize the Mandelbrot set to higher dimensions (see [1], [2], and [13]), and in theoretical physics to generalize the linear and non-linear Schrödinger’s equation [25]. The second section of this paper gives some preliminaries on the multicomplex numbers.

We note that counting involutions is very natural in many other settings than real algebras, where involutions play a fundamental role. Perhaps the most prominent example is counting involutions on the symmetric group $S_{n}$ (see, for instance, [6, 18]). Related to this are signed permutations and, in particular, signed involutions. A signed permutation of length $n$ is a permutation of $\{1,2,\ldots,n\}$ written in one-line notation where each entry may have a bar over it. For instance, $\pi=3\overline{1}2$ is a signed permutation. We write $B_{n}$ for the set of signed permutations of length $n$ , which also corresponds to the group of symmetries of a hypercube, the hyperoctahedral group, which is a Coxeter group of type $B$ and of rank $n$ [5, 11, 22]. The first main result of this paper establishes a connection between real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ and $B_{2^{n-1}}$ .

Theorem 1.1.

For each integer $n\geq 1$ , there is a bijection between the set of real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ and $B_{2^{n-1}}$ . Furthermore, this bijection sends the identity function to the identity signed permutation and is compatible with composition.

The proof of this theorem is given in Section 3. The idea of the proof is based on the idempotent representation of a multicomplex number and on a representation theorem for the numbers squaring to $\pm 1$ . In particular, we prove in section 2 the following elementary fact pertaining to multicomplex numbers, which was surprisingly unknown in the literature.

Proposition 1.2.

For each integer $n\geq 1$ , there are $2^{2^{n-1}}$ multicomplex numbers squaring to $1$ , and $2^{2^{n-1}}$ squaring to $-1$ .

We also prove that the same is true for idempotent elements, which resolves a recent conjecture [28].

Proposition 1.3.

For each integer $n\geq 1$ , there are $2^{2^{n-1}}$ idempotent elements in $\mathbb{M}\mathbb{C}(n)$ .

We first use Theorem 1.1 to count the number of real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ .

Corollary 1.4.

For each integer $n\geq 1$ , there are

2^{2^{n-1}}(2^{n-1})!

real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ .

It also allows us to obtain a formula giving the number of involutions of the multicomplex numbers, which was our goal, and to deduce a few more corollaries.

Corollary 1.5.

For $n\geq 1$ a positive integer, write $F(n)$ for the number of involutions of $\mathbb{M}\mathbb{C}(n)$ .

(i)

The following formula holds:

$F(n)=(2^{n-1})!\sum_{k=0}^{\lfloor 2^{n-2}\rfloor}\frac{2^{2^{n-1}-2k}}{k!(2^{n-1}-2k)!}.$
(ii)

If $g(1)=2$ , $g(2)=6$ , and

$\displaystyle g(n)=2g(n-1)+(2n-2)g(n-2),\qquad n\geq 3,$

then $F(n)=g(2^{n-1})$ .
(iii)

The asymptotics

$F(n)\sim\Bigl{(}\frac{2^{n}}{e}\Bigr{)}^{2^{n-2}}\frac{e^{2^{n/2}}}{\sqrt{2e}}$

hold as $n\to\infty$ .

Note that $g(n)$ is documented on the OEIS under sequence A000898 [19], where more results on $g(n)$ (hence on $F(n)$ ) can be found. Also, considering the first few values of $F(n)$ as in the following table tells us that the fast growth rate of the previous asymptotics is not surprising.

$n$	1	2	3	4	5
$F(n)$	2	6	76	32,400	50,305,536,256

Remark.

Theorem 1.1 could be used to deduce more results on certain subsets of real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ from known results on $B_{n}$ , but this would be rather artificial as these results would not necessarily translate mutatis mutandis in the language of multicomplex functions. However, studying involutions on multicomplex spaces via our bijection is natural because the involutions of $\mathbb{M}\mathbb{C}(n)$ precisely correspond to signed involutions on $B_{2^{n-1}}$ .

We also study $r$ -involutions of the multicomplex numbers. We call a function $f$ of $\mathbb{M}\mathbb{C}(n)$ an $r$ -involution if $f$ is a real-linear ring homomorphism such that $f^{(r)}=\mathrm{Id}$ , where $f^{(r)}$ is the $r$ -th composition of $f$ and $\mathrm{Id}$ is the identity map. Note that from this definition, it follows that if $f$ is an $r$ -involution, then it is also an $mr$ -involution for any positive integer $m$ .

We denote by $F_{r}(n)$ the number of such $r$ -involutions on $\mathbb{M}\mathbb{C}(n)$ . We define $S_{n,t}$ as the set of permutations $\sigma$ of $n$ elements such that $\sigma^{(t)}=\mathrm{Id}.$ For a given permutation $\sigma$ , we denote by $\mathrm{cyc}_{k}(\sigma)$ the number of disjoint cycles of length $k$ in $\sigma$ . Using Theorem 1.1 again, we prove in section 4 the following more general result

Corollary 1.6.

Let $p>2$ be a prime number. The number of $p$ -involutions of the multicomplex numbers $\mathbb{MC}(n)$ with $n\geq 1$ is given by

\displaystyle F_{p}(n)=(2^{n-1})!\sum_{k=0}^{\lfloor 2^{n-1}/p\rfloor}\frac{2^{(p-1)k}}{k!p^{k}(2^{n-1}-pk)!}.

More generally, if $r>1$ is a positive integer, the number of $r$ -involutions of the multicomplex numbers $\mathbb{MC}(n)$ with $n\geq 1$ is given by

\displaystyle F_{r}(n)=2^{2^{n-1}}\sum_{\sigma\in S_{2^{n-1},r}}\Big{(}\prod_{k|r,\,r/k\mathrm{\,is\,odd}}\frac{1}{2^{\mathrm{cyc}_{k}(\sigma)}}\Big{)}.

Remark that if $p>2^{n-1}$ , then $F_{p}(n)=1$ , and the only counted $p$ -involution is the identity map. Also, in the second part of Corollary 1.6, it is clear that $\mathrm{cyc}_{k}(\sigma)=0$ if $k>2^{n-1}$ . Hence, if $r>2^{n-1}$ , the product could be restricted to the values of $k$ such that $k\leq 2^{n-1}$ .

The previous Corollary and Theorem 1.1 imply the following result on signed $r$ -involutions of $B_{n}$ , which we could not find in the literature, although it could be deduced from stronger results on $r$ -involutions in more general groups (see, e.g., [4]).

Corollary 1.7.

Fix $n\geq 1$ a positive integer. For $r>1$ , the number of signed $r$ -involutions on $B_{n}$ is

2^{n}\sum_{\sigma\in S_{n,r}}\Big{(}\prod_{k|r,\,r/k\mathrm{\,is\,odd}}\frac{1}{2^{\mathrm{cyc}_{k}(\sigma)}}\Big{)}.

In particular, for $p>2$ a prime number, the number of $p$ -involutions on $B_{n}$ is

n!\sum_{k=0}^{\lfloor n/p\rfloor}\frac{2^{(p-1)k}}{k!p^{k}(n-pk)!}.

Our last main result gives a count for a particular type of involutions on $\mathbb{M}\mathbb{C}(n)$ . We let $\mathbf{i_{1}},\ldots,\mathbf{i_{n}}$ be the elementary commuting imaginary units of $\mathbb{MC}(n)$ and define the set $\mathbb{I}(n)$ as the set of numbers that can be written as $\mathbf{i_{1}}^{a_{1}}\cdots\mathbf{i_{n}}^{a_{n}}$ with $a_{k}\in\{0,1\}.$ Observe that since $\mathbf{i_{k}^{2}}=-1,\,1\leq k\leq n$ , and the elementary units commute, the elements of $\mathbb{I}(n)$ square to $\pm 1$ . For this introduction, it is sufficient to mention that a multicomplex number $\eta$ can be written as a linear combination of these units as

\displaystyle\eta=\sum_{\mathbf{i}\in\mathbb{I}(n)}\eta_{\mathbf{i}}\mathbf{i},

where $\eta_{\mathbf{i}}\in\mathbb{R}$ . The proof of Proposition 1.2 highlighted a surprising phenomenon in the multicomplex numbers: there are numbers squaring to $-1$ and $1$ that are not in the set of units $\mathbb{I}(n)$ . Our observation motivates the following question: how many involutions of $\mathbb{M}\mathbb{C}(n)$ send the units of $\mathbb{I}(n)$ to the units of $\mathbb{I}(n)$ ? We call such involutions $\mathbb{I}(n)$ -preserving involutions. The following theorem gives an answer.

Theorem 1.8.

The number of $\mathbb{I}(n)$ -preserving involutions of $\mathbb{M}\mathbb{C}(n)$ , with $n\geq 1$ , is

\displaystyle\sum_{k=\lceil n/2\rceil}^{n}\Big{(}\prod_{j=1}^{k-1}\frac{2^{n}-2^{j}}{2^{k}-2^{j}}\Big{)}\Big{(}\prod_{j=0}^{n-k-1}(2^{k}-2^{j})\Big{)}2^{k},

where an empty product is understood to be equal to $1$ .

The proof of Theorem 1.8 relies on completely different tools. The main idea is to transfer the problem of counting involutions on $\mathbb{M}\mathbb{C}(n)$ to counting matrices with entries in $\{0,1\}$ . The proof of the last theorem is presented in Section 5. We note that we could not find a reference for the sequence of values given by the formula in Theorem 1.8 on the On-Line Encyclopedia of Integer Sequences. Moreover, Theorem 1.8 is a more natural generalization of the analogue result for bicomplex numbers from [20]. Indeed, in the bicomplex numbers, the involutions preserve the set of units $\mathbb{I}(2)$ .

2. Background on multicomplex numbers

Multicomplex numbers were introduced by Segre [23] and Cockle [7, 8, 9, 10] to give a generalization of the complex numbers to $\mathbb{C}^{n}$ , for $n\geq 2$ . For a modern treatment of the multicomplex numbers, we refer the reader to [21]. We will mainly follow the presentation given in [2], with some changes in the notation.

2.1. Multicomplex numbers

The definition of the multicomplex numbers is given recursively. Let $\mathbb{M}\mathbb{C}(0)$ be the set of real numbers and let $\mathbb{M}\mathbb{C}(n)$ , $n\geq 1$ , be the set

(2.1)

\displaystyle\mathbb{M}\mathbb{C}(n):=\{\eta=\eta_{1}+\eta_{2}\mathbf{i_{n}}\,:\,\eta_{1},\eta_{2}\in\mathbb{M}\mathbb{C}(n-1),\,\mathbf{i_{n}^{2}}=-1\}.

For example, when $n=1$ , we obtain the set $\mathbb{M}\mathbb{C}(1)$ of complex numbers $\eta_{1}+\eta_{2}\mathbf{i_{1}}$ , where $\mathbf{i_{1}^{2}}=-1$ . When $n=2$ , we obtain the set $\mathbb{M}\mathbb{C}(2)$ of bicomplex numbers $\eta_{1}+\eta_{2}\mathbf{i_{2}}$ , where $\eta_{1},\eta_{2}$ are complex numbers, $\mathbf{i_{2}^{2}}=-1$ , and $\mathbf{i_{1}}\neq\mathbf{i_{2}}$ . We say that two multicomplex numbers $\eta$ and $\zeta$ are equal if and only if $\eta_{1}=\zeta_{1}$ and $\eta_{2}=\zeta_{2}$ . If we let $\eta_{2}=0$ in the expression of a multicomplex number $\eta=\eta_{1}+\eta_{2}\mathbf{i_{n}}$ , we see that $\mathbb{M}\mathbb{C}(n-1)\subset\mathbb{M}\mathbb{C}(n)$ .

The set of multicomplex numbers becomes a commutative ring if we endow it with the following algebraic operations:

•

$\eta+\zeta:=(\eta_{1}+\zeta_{1})+(\eta_{2}+\zeta_{2})\mathbf{i_{n}}$ ;
•

$\eta\zeta:=(\eta_{1}\zeta_{1}-\eta_{2}\zeta_{2})+(\eta_{1}\zeta_{2}+\eta_{2}\zeta_{1})\mathbf{i_{n}}$ .

These last operations must be understood recursively.

Let $\eta=\eta_{1}+\eta_{2}\mathbf{i_{n}}$ be a multicomplex number. This means that $\eta_{1},\eta_{2}\in\mathbb{M}\mathbb{C}(n-1)$ . Therefore, there are multicomplex numbers $\eta_{11},\eta_{12},\eta_{21},\eta_{22}\in\mathbb{M}\mathbb{C}(n-2)$ such that $\eta_{1}=\eta_{11}+\eta_{12}\mathbf{i_{n-1}}$ and $\eta_{2}=\eta_{21}+\eta_{22}\mathbf{i_{n-1}}$ . Replacing the $\eta_{1}$ and $\eta_{2}$ in the expression for $\eta$ , we obtain the representation of a multicomplex number in $\mathbb{M}\mathbb{C}(n)$ in terms of four components in $\mathbb{M}\mathbb{C}(n-2)$ :

\displaystyle\zeta=(\eta_{11}+\eta_{12}\mathbf{i_{n-1}})+(\eta_{21}+\eta_{22}\mathbf{i_{n-1}})\mathbf{i_{n}}.

From the definition of the multiplication, we can distribute $\mathbf{i_{n}}$ to obtain

\displaystyle\zeta=\eta_{11}+\eta_{12}\mathbf{i_{n-1}}+\eta_{21}\mathbf{i_{n}}+\eta_{22}\mathbf{i_{n-1}}\mathbf{i_{n}}.

For example, a bicomplex number $\eta=\eta_{1}+\eta_{2}\mathbf{i_{2}}$ can be expressed in a linear combination of four real numbers as follows:

\displaystyle\eta=\eta_{11}+\eta_{12}\mathbf{i_{1}}+\eta_{21}\mathbf{i_{2}}+\eta_{22}\mathbf{i_{1}}\mathbf{i_{2}}.

We can continue this process recursively until we reach the set $\mathbb{M}\mathbb{C}(0)$ . At each stage $k$ ( $1\leq k\leq n$ ) of the process, we obtain a representation of a multicomplex number in terms of $2^{k}$ multicomplex numbers in $\mathbb{M}\mathbb{C}(n-k)$ . All of these representations are called the canonical representation (or the cartesian representation) of a multicomplex number. The canonical representation we are interested in is the one in terms of $2^{n}$ components in $\mathbb{M}\mathbb{C}(0)$ . To be more explicit, recall that $\mathbb{I}(n)$ is the set of all different possible products of the elements in the set $\{1,\mathbf{i_{1}},\mathbf{i_{2}},\ldots,\mathbf{i_{n}}\}$ . Since the multiplication is commutative, the cardinality of $\mathbb{I}(n)$ is $2^{n}$ . Therefore, we can write any multicomplex number as

(2.2)

\displaystyle\eta=\sum_{\mathbf{i}\in\mathbb{I}(n)}\eta_{\mathbf{i}}\mathbf{i},

where $\eta_{\mathbf{i}}\in\mathbb{R}$ . This tells us that the elements of $\mathbb{I}(n)$ form a basis of $\mathbb{M}\mathbb{C}(n)$ . For instance, when $n=2$ or $3$ , the following holds:

•

For $\eta\in\mathbb{M}\mathbb{C}(2)$ , we have

\displaystyle\eta=\eta_{1}+\eta_{\mathbf{i_{1}}}\mathbf{i_{1}}+\eta_{\mathbf{i_{2}}}\mathbf{i_{2}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}\mathbf{i_{1}}\mathbf{i_{2}}.

•

For $\eta\in\mathbb{M}\mathbb{C}(3)$ , we have

\displaystyle\eta=\eta_{1}+\eta_{\mathbf{i_{1}}}\mathbf{i_{1}}+\eta_{\mathbf{i_{2}}}\mathbf{i_{2}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}\mathbf{i_{1}}\mathbf{i_{2}}+\eta_{\mathbf{i_{3}}}\mathbf{i_{3}}+\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}\mathbf{i_{1}}\mathbf{i_{3}}+\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{2}}\mathbf{i_{3}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}.

Using this representation and the algebraic operations defined above, we can view the set $\mathbb{M}\mathbb{C}(n)$ as a commutative and associative algebra on the set of real numbers.

2.2. An idempotent representation for multicomplex numbers

Of particular importance in the set of multicomplex numbers are the numbers $\eta$ such that $\eta^{2}=\eta$ , which are called idempotent numbers. In particular, we consider

\displaystyle\mathbf{e_{n}}:=\frac{1+\mathbf{i_{n-1}}\mathbf{i_{n}}}{2}\quad\text{and}\quad\mathbf{\overline{e}_{n}}:=\frac{1-\mathbf{i_{n-1}}\mathbf{i_{n}}}{2}.

An additional property that these numbers have is that $\mathbf{e_{n}}\mathbf{\overline{e}_{n}}=0$ . If we multiply a multicomplex number $\eta=\eta_{1}+\eta_{2}\mathbf{i_{n}}$ by $\mathbf{e_{n}}$ and by $\mathbf{\overline{e}_{n}}$ respectively, we obtain

\displaystyle\eta\mathbf{e_{n}}=(\eta_{1}-\eta_{2}\mathbf{i_{n-1}})\mathbf{e_{n}}\quad\text{and}\quad\eta\mathbf{\overline{e}_{n}}=(\eta_{1}+\eta_{2}\mathbf{i_{n-1}})\mathbf{\overline{e}_{n}}.

Since $\mathbf{e_{n}}+\mathbf{\overline{e}_{n}}=1$ , summing $\eta\mathbf{e_{n}}$ and $\eta\mathbf{\overline{e}_{n}}$ yields the idempotent representation of a multicomplex number, namely

\displaystyle\eta=(\eta_{1}-\eta_{2}\mathbf{i_{n-1}})\mathbf{e_{n}}+(\eta_{1}+\eta_{2}\mathbf{i_{n-1}})\mathbf{\overline{e}_{n}}.

We see that the numbers multiplying $\mathbf{e_{n}}$ and $\mathbf{\overline{e}_{n}}$ are elements of $\mathbb{M}\mathbb{C}(n-1)$ , which we call the idempotent components of $\eta$ . We will denote them by $\eta_{\mathbf{e_{n}}}$ and $\eta_{\mathbf{\overline{e}_{n}}}$ , respectively. The idempotent representation can therefore be rewritten as

(2.3)

\displaystyle\eta=\eta_{\mathbf{e_{n}}}\mathbf{e_{n}}+\eta_{\mathbf{\overline{e}_{n}}}\mathbf{\overline{e}_{n}}.

Note that two multicomplex numbers are equal if and only if their idempotent components are equal.

The idempotent representation is important because it transforms the multiplication of multicomplex numbers into a component-wise multiplication. More precisely, if $\eta=\eta_{\mathbf{e_{n}}}\mathbf{e_{n}}+\eta_{\mathbf{\overline{e}_{n}}}\mathbf{\overline{e}_{n}}$ and $\zeta=\zeta_{\mathbf{e_{n}}}\mathbf{e_{n}}+\zeta_{\mathbf{\overline{e}_{n}}}\mathbf{\overline{e}_{n}}$ , then we have

(2.4)

\displaystyle\eta\zeta=\eta_{\mathbf{e_{n}}}\zeta_{\mathbf{e_{n}}}\mathbf{e_{n}}+\eta_{\mathbf{\overline{e}_{n}}}\zeta_{\mathbf{\overline{e}_{n}}}\mathbf{\overline{e}_{n}}.

We now apply this result to the idempotent components of a multicomplex number $\eta$ . Define

\displaystyle\mathbf{e_{n-1}}:=\frac{1+\mathbf{i_{n-2}}\mathbf{i_{n-1}}}{2}\quad\text{and}\quad\mathbf{\overline{e}_{n-1}}:=\frac{1-\mathbf{i_{n-2}}\mathbf{i_{n-1}}}{2}.

Then, the idempotent components $\eta_{\mathbf{e_{n}}}$ and $\eta_{\mathbf{\overline{e}_{n}}}$ of $\eta\in\mathbb{M}\mathbb{C}(n)$ can be written as

\displaystyle\eta_{\mathbf{e_{n}}}=\eta_{\mathbf{e_{n-1}}\mathbf{e_{n}}}\mathbf{e_{n-1}}+\eta_{\mathbf{\overline{e}_{n-1}}\mathbf{e_{n}}}\mathbf{\overline{e}_{n-1}}

and

\displaystyle\eta_{\mathbf{\overline{e}_{n}}}=\eta_{\mathbf{e_{n-1}}\mathbf{\overline{e}_{n}}}\mathbf{e_{n-1}}+\eta_{\mathbf{\overline{e}_{n-1}}\mathbf{\overline{e}_{n}}}\mathbf{\overline{e}_{n-1}},

where $\eta_{\mathbf{e_{n-1}}\mathbf{e_{n}}},\eta_{\mathbf{\overline{e}_{n-1}}\mathbf{e_{n}}},\eta_{\mathbf{e_{n-1}}\mathbf{\overline{e}_{n}}},\eta_{\mathbf{\overline{e}_{n-1}}\mathbf{\overline{e}_{n}}}\in\mathbb{M}\mathbb{C}(n-2)$ . Replacing these in the idempotent representation of $\eta\in\mathbb{M}\mathbb{C}(n)$ , we obtain a second idempotent representation in terms of components in $\mathbb{M}\mathbb{C}(n-2)$ :

\displaystyle\eta=\eta_{\mathbf{e_{n-1}}\mathbf{e_{n}}}\mathbf{e_{n-1}}\mathbf{e_{n}}+\eta_{\mathbf{\overline{e}_{n-1}}\mathbf{e_{n}}}\mathbf{\overline{e}_{n-1}}\mathbf{e_{n}}+\eta_{\mathbf{e_{n-1}}\mathbf{\overline{e}_{n}}}\mathbf{e_{n-1}}\mathbf{\overline{e}_{n}}+\eta_{\mathbf{\overline{e}_{n-1}}\mathbf{\overline{e}_{n}}}\mathbf{\overline{e}_{n-1}}\mathbf{\overline{e}_{n}}.

More generally, define the following elements for each integer $k\geq 2$ :

\displaystyle\mathbf{e_{k}}:=\frac{1+\mathbf{i_{k-1}}\mathbf{i_{k}}}{2}\quad\text{and}\quad\mathbf{\overline{e}_{k}}:=\frac{1-\mathbf{i_{k-1}}\mathbf{i_{k}}}{2}.

We then define a family of sets $\mathcal{E}(k,n)$ inductively for $n\geq 2$ and $2\leq k\leq n$ :

•

$\mathcal{E}(n,n):=\{\mathbf{e_{n}},\mathbf{\overline{e}_{n}}\}$ for $k=n$ ;
•

$\mathcal{E}(k,n):=\mathcal{E}(k+1,n)\mathbf{e_{k}}\cup\mathcal{E}(k+1,n)\mathbf{\overline{e}_{k}}$ for $2\leq k<n$ .

Now, for any $2\leq k\leq n$ , an induction argument shows that the cardinality of $\mathcal{E}(k,n)$ is $2^{n-k+1}$ . Also, by induction, we have that if $\bm{\varepsilon}\in\mathcal{E}(k,n)$ , then $\bm{\varepsilon}^{2}=\bm{\varepsilon}$ , and if ${\bm{\varepsilon}}_{1},{\bm{\varepsilon}}_{2}\in\mathcal{E}(k,n)$ with ${\bm{\varepsilon}}_{1}\neq{\bm{\varepsilon}}_{2}$ , then ${\bm{\varepsilon}}_{1}{\bm{\varepsilon}}_{2}=0$ .

Finally, any multicomplex number $\eta\in\mathbb{M}\mathbb{C}(n)$ can be rewritten as

\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}(k,n)}\eta_{\bm{\varepsilon}}\bm{\varepsilon},

where $\eta_{\bm{\varepsilon}}\in\mathbb{M}\mathbb{C}(k-1)$ for all $\bm{\varepsilon}\in\mathcal{E}(k,n)$ .

These new idempotent representations still have the advantage of simplifying the operation of multiplication. If

\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}(k,n)}\eta_{\bm{\varepsilon}}\bm{\varepsilon}\quad\text{and}\quad\zeta=\sum_{\bm{\varepsilon}\in\mathcal{E}(k,n)}\zeta_{\bm{\varepsilon}}\bm{\varepsilon},

then the following holds:

•

$\eta=\zeta$ if and only if $\eta_{\bm{\varepsilon}}=\zeta_{\bm{\varepsilon}}$ for all $\bm{\varepsilon}\in\mathcal{E}(k,n)$ ;
•

$\eta+\zeta=\sum_{\bm{\varepsilon}\in\mathcal{E}(k,n)}(\eta_{\bm{\varepsilon}}+\zeta_{\bm{\varepsilon}})\bm{\varepsilon}$ ;
•

$\eta\zeta=\sum_{\bm{\varepsilon}\in\mathcal{E}(k,n)}(\eta_{\bm{\varepsilon}}\zeta_{\bm{\varepsilon}})\bm{\varepsilon}$ .

There is a special case we will use later on in this paper. We denote the set $\mathcal{E}(2,n)$ by $\mathcal{E}_{n}$ . We can write any multicomplex number $\eta$ as

(2.5)

\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\eta_{\bm{\varepsilon}}\bm{\varepsilon},

where $\eta_{\bm{\varepsilon}}\in\mathbb{M}\mathbb{C}(1)$ for all $\bm{\varepsilon}\in\mathcal{E}_{n}$ .

2.3. Representation theorems and bijections

We use the notation $U_{n}$ to denote the set of multicomplex numbers squaring to $-1$ and $H_{n}$ for numbers squaring to $1$ , namely

\displaystyle U_{n}:=\{\eta\in\mathbb{M}\mathbb{C}(n)\,:\,\eta^{2}=-1\}\quad\text{and}\quad H_{n}:=\{\eta\in\mathbb{M}\mathbb{C}(n)\,:\,\eta^{2}=1\}.

We also write $E_{n}$ for the set of idempotent elements of $\mathbb{M}\mathbb{C}(n)$ , that is,

E_{n}:=\{\eta\in\mathbb{M}\mathbb{C}(n)\,:\,\eta^{2}=\eta\}.

For example, $U_{3}$ contains the numbers

(1)

$\mathbf{i_{1}}$ , $-\mathbf{i_{1}}$ ;
(2)

$\mathbf{i_{2}}$ , $-\mathbf{i_{2}}$ ;
(3)

$\mathbf{i_{3}}$ , $-\mathbf{i_{3}}$ ;
(4)

$\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}$ , $-\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}$ ;
(5)

$\frac{\mathbf{i_{1}}+\mathbf{i_{2}}+\mathbf{i_{3}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{\mathbf{i_{1}}+\mathbf{i_{2}}+\mathbf{i_{3}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(6)

$\frac{\mathbf{i_{1}}-\mathbf{i_{2}}-\mathbf{i_{3}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{\mathbf{i_{1}}-\mathbf{i_{2}}-\mathbf{i_{3}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(7)

$\frac{\mathbf{i_{1}}+\mathbf{i_{2}}-\mathbf{i_{3}}-\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{\mathbf{i_{1}}+\mathbf{i_{2}}-\mathbf{i_{3}}-\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(8)

$\frac{\mathbf{i_{1}}-\mathbf{i_{2}}+\mathbf{i_{3}}-\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{\mathbf{i_{1}}-\mathbf{i_{2}}+\mathbf{i_{3}}-\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ,

while $H_{3}$ contains

(1)

$1$ , $-1$ ;
(2)

$\mathbf{i_{1}}\mathbf{i_{2}}$ , $-\mathbf{i_{1}}\mathbf{i_{2}}$ ;
(3)

$\mathbf{i_{1}}\mathbf{i_{3}}$ , $-\mathbf{i_{1}}\mathbf{i_{3}}$ ;
(4)

$\mathbf{i_{2}}\mathbf{i_{3}}$ , $-\mathbf{i_{2}}\mathbf{i_{3}}$ ;
(5)

$\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(6)

$\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(7)

$\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(8)

$\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $-\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ,

and $E_{3}$ contains

(1)

$1$ , $0$ ;
(2)

$\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}}{2}$ , $\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}}{2}$ ;
(3)

$\frac{1+\mathbf{i_{1}}\mathbf{i_{3}}}{2}$ , $\frac{1-\mathbf{i_{1}}\mathbf{i_{3}}}{2}$ ;
(4)

$\frac{1+\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ , $\frac{1-\mathbf{i_{2}}\mathbf{i_{3}}}{2}$ ;
(5)

$\frac{3+\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ , $\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ ;
(6)

$\frac{3-\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ , $\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ ;
(7)

$\frac{3+\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ , $\frac{1-\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ ;
(8)

$\frac{3-\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ , $\frac{1+\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}}}{4}$ .

Also note that the set $E_{n}$ contains many more elements than the set $\mathcal{E}_{n}$ . For example, the elements of $\mathcal{E}_{3}$ correspond to the second number in items (5)-(8) in the list of elements of $E_{3}$ .

The above examples and Proposition 1.2 suggest that there is a bijection between $U_{n}$ and $H_{n}$ . It can be given explicitly as follows.

Proposition 2.1.

Let $u$ be any element of $U_{n}$ . We have

U_{n}=uH_{n}\quad\text{and}\quad H_{n}=uU_{n}.

Proof.

Fix $u\in U_{n}$ . Define the map $f:H_{n}\rightarrow U_{n}$ by $f(h):=uh$ . Then $f$ is injective. If $v\in U_{n}$ , then $-uv\in H_{n}$ because $(-uv)^{2}=1$ . Therefore, there is an $h\in H_{n}$ such that $h=-uv$ and so $uh=v$ . This means that $f$ is surjective and therefore $f$ is a bijection. Since $f(H_{n})=U_{n}$ , we obtain the first equality. The second equality is obtained similarly. ∎

We are now ready to prove Proposition 1.2.

Proof of Proposition 1.2.

By Proposition 2.1, it suffices to prove it for $U_{n}$ . We proceed by induction on $n$ . For $n=1$ , the set $\mathbb{M}\mathbb{C}(1)$ is the set of complex numbers and there are only two solutions to $\eta^{2}=-1$ , namely $\mathbf{i_{1}}$ and $-\mathbf{i_{1}}$ . This is exactly $2^{2^{n-1}}$ for $n=1$ . Suppose that there are $2^{2^{n-1}}$ solutions for $\eta^{2}=-1$ in $\mathbb{M}\mathbb{C}(n)$ . We will show that there are $2^{2^{n}}$ solutions for $\eta^{2}=-1$ in $\mathbb{M}\mathbb{C}(n+1)$ . Let $\eta\in\mathbb{M}\mathbb{C}(n+1)$ be written in its idempotent representation, that is, $\eta=\eta_{\mathbf{e_{n+1}}}\mathbf{e_{n+1}}+\eta_{\mathbf{\overline{e}_{n+1}}}\mathbf{\overline{e}_{n+1}}$ , where $\eta_{\mathbf{e_{n+1}}},\eta_{\mathbf{\overline{e}_{n+1}}}\in\mathbb{M}\mathbb{C}(n)$ . Then, according to (2.4), $\eta$ is a solution to the equation $\eta^{2}=-1$ if and only if $(\eta_{\mathbf{e_{n+1}}},\eta_{\mathbf{\overline{e}_{n+1}}})$ is a solution to the system of equations

\displaystyle\left\{\begin{matrix}\eta_{\mathbf{e_{n+1}}}^{2}=-1,\\ \eta_{\mathbf{\overline{e}_{n+1}}}^{2}=-1.\end{matrix}\right.

Since $\eta_{\mathbf{e_{n+1}}},\eta_{\mathbf{\overline{e}_{n+1}}}\in\mathbb{M}\mathbb{C}(n)$ , by the induction hypothesis there are $2^{2^{n-1}}$ solutions to each equation in the system. Therefore, there are $2^{2^{n-1}}\cdot 2^{2^{n-1}}=2^{2^{n}}$ solutions to $\eta^{2}=-1$ in $\mathbb{M}\mathbb{C}(n+1)$ . This ends the induction and the claim is proved. ∎

To prove Proposition 1.3, one could modify the proof of Proposition 1.2 according to $\eta^{2}=\eta$ . We prefer to give an explicit bijection between $H_{n}$ and $E_{n}$ .

Proposition 2.2.

The map $f:H_{n}\to E_{n}$ defined by $f(h)=(1+h)/2$ is a bijection.

Proof.

If $h\in H_{n}$ , then

\Big{(}\frac{1+h}{2}\Big{)}^{2}=\frac{1+h}{2}.

It follows that the map $h\mapsto(1+h)/2$ is well defined. It is also clearly a bijection. ∎

An interesting corollary to the proofs of Proposition 1.2 and Proposition 1.3 is the following representation theorem for certain multicomplex numbers.

Corollary 2.3.

Let $\eta\in\mathbb{M}\mathbb{C}(n)$ .

(i)

If $\eta^{2}=-1$ , then $\eta$ can be written as

$\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\eta_{\bm{\varepsilon}}\bm{\varepsilon}$

where the components $\eta_{\bm{\varepsilon}}\in\{\mathbf{i_{1}},-\mathbf{i_{1}}\}$ .
(ii)

If $\eta^{2}=1$ , then $\eta$ can be written as

$\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\eta_{\bm{\varepsilon}}\bm{\varepsilon}$

where the components $\eta_{\bm{\varepsilon}}\in\{1,-1\}$ .
(iii)

If $\eta^{2}=\eta$ , then $\eta$ can be written as

$\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\eta_{\bm{\varepsilon}}\bm{\varepsilon}$

where the components $\eta_{\bm{\varepsilon}}\in\{0,1\}$ .

Proof.

The result for numbers squaring to $-1$ is immediate. Then (ii) follows from Proposition 2.1. Part (iii) follows from Proposition 2.2 and the fact that $1=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\bm{\varepsilon}$ . ∎

3. Characterization of Involutions of $\mathbb{M}\mathbb{C}(n)$

We give a precise definition of what we mean by an involution on the set $\mathbb{M}\mathbb{C}(n)$ .

Definition 3.1.

A function $f:\mathbb{M}\mathbb{C}(n)\rightarrow\mathbb{M}\mathbb{C}(n)$ is said to be an involution if the following conditions are satisfied:

(1)

$f(f(\eta))=\eta$ for any $\eta\in\mathbb{M}\mathbb{C}(n)$ ;
(2)

$f(\eta+\zeta)=f(\eta)+f(\zeta)$ and $f(\lambda\eta)=\lambda f(\eta)$ for any $\eta,\zeta\in\mathbb{M}\mathbb{C}(n)$ and $\lambda\in\mathbb{R}$ ;
(3)

$f(\eta\zeta)=f(\eta)f(\zeta)$ for any $\eta,\zeta\in\mathbb{M}\mathbb{C}(n)$ .

The usual definition of an involution involves only the first condition. However, for the quaternions, bicomplex numbers and general algebras over commutative fields (see [12], [14], [20]), the above definition was adopted. Therefore, to be consistent with these references, we will stick to the above definition. If we take a closer look at our definition of the term “involution”, we require that the function is a real-linear homomorphism which is its own inverse. When we do not require that $f$ be its own inverse, we shall only say that $f$ is a real-linear automorphism of $\mathbb{M}\mathbb{C}(n)$ .

3.1. Auxiliary results

We now turn our attention to Theorem 1.1. Recall that for any $\eta\in\mathbb{M}\mathbb{C}(n)$ , we can write

\displaystyle\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\eta_{\bm{\varepsilon}}\bm{\varepsilon},

where $\eta_{\bm{\varepsilon}}\in\mathbb{M}\mathbb{C}(1)$ . Write $\eta_{\bm{\varepsilon}}=x_{\bm{\varepsilon}}+\mathbf{i_{1}}y_{\bm{\varepsilon}}$ with $x_{\bm{\varepsilon}},y_{\bm{\varepsilon}}\in\mathbb{R}$ . Therefore, for $f:\mathbb{M}\mathbb{C}(n)\rightarrow\mathbb{M}\mathbb{C}(n)$ a real-linear ring homomorphism, we have

\displaystyle f(\eta)=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}x_{\bm{\varepsilon}}f(\bm{\varepsilon})+f(\mathbf{i_{1}})\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}y_{\bm{\varepsilon}}f(\bm{\varepsilon}).

As an immediate consequence, we obtain the following proposition.

Proposition 3.2.

An involution of $\mathbb{M}\mathbb{C}(n)$ is completely determined by its value on $\mathbf{i_{1}}$ and on the set $\mathcal{E}_{n}$ .

To choose the value $f(\mathbf{i_{1}})$ , a key ingredient is the following observation.

Corollary 3.3.

Let $f$ be an involution of $\mathbb{M}\mathbb{C}(n)$ . Then the following assertions hold.

(i)

Given a multicomplex number $\eta$ such that $\eta^{2}=-1$ , there exists a unique $h\in H_{n}$ such that $\eta=\mathbf{i_{1}}h$ .
(ii)

$f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ for a choice of $h\in H_{n}$ that depends on $f$ .

Proof.

To prove (i), apply Proposition 2.1 with $u=\mathbf{i_{1}}$ . Part (ii) follows from (i) and the fact that $f(\mathbf{i_{1}})^{2}=-1$ . ∎

Now we shall see how $f$ acts on $\mathcal{E}_{n}$ . Let $\eta$ be an element of $E_{n}$ . Define the set

\mathrm{orth}(\eta)=\{\zeta:\zeta^{2}=\zeta,\,\eta\zeta=0\},

that is, the set of idempotent elements orthogonal to $\eta$ . We write

(3.1)

\eta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\eta_{\bm{\varepsilon}}\bm{\varepsilon}

and $v(\eta)$ for the number of coefficients equal to 0 in the right-hand side of (3.1). If the number $\zeta$ is such that $\zeta^{2}=\zeta$ and $\zeta=\sum_{\bm{\varepsilon}\in\mathcal{E}_{n}}\zeta_{\bm{\varepsilon}}\bm{\varepsilon}$ , then the equality $\zeta\eta=0$ is equivalent to $\eta_{\bm{\varepsilon}}\zeta_{\bm{\varepsilon}}=0$ for all $\bm{\varepsilon}\in\mathcal{E}_{n}$ . If $\eta_{\bm{\varepsilon}}=0$ , $\zeta_{\bm{\varepsilon}}$ can take the values 0 or 1, while if $\eta_{\bm{\varepsilon}}=1$ , $\zeta_{\bm{\varepsilon}}$ must be equal to 0. We thus have

•

$\#\mathrm{orth}(\eta)=2^{v(\eta)}$ ;
•

$\#\mathrm{orth}(0)=2^{2^{n-1}}$ ;
•

If $\bm{\varepsilon}\in\mathcal{E}_{n}$ , then $\#\mathrm{orth}(\bm{\varepsilon})=2^{2^{n-1}-1};$
•

If $\eta\neq 0$ and $\eta\not\in\mathcal{E}_{n}$ , then $\#\mathrm{orth}(\eta)<2^{2^{n-1}-1}$ .

In the above statements, the notation $\#A$ means the cardinality of the set $A$ . A second key ingredient in proving our main result is the following lemma describing the action of $f$ on the element of $\mathcal{E}_{n}$ .

Lemma 3.4.

Let $f$ be a bijection from $\mathbb{M}\mathbb{C}(n)$ to $\mathbb{M}\mathbb{C}(n)$ such that $f(0)=0$ and $f(\eta\zeta)=f(\eta)f(\zeta)$ . Let $\eta$ be such that $\eta^{2}=\eta$ . Then

(i)

$(f(\eta))^{2}=f(\eta)$ .
(ii)

$\#\mathrm{orth}(f(\eta))=\#\mathrm{orth}(\eta)$ .
(iii)

If $\bm{\varepsilon}\in\mathcal{E}_{n}$ , then $f(\bm{\varepsilon})\in\mathcal{E}_{n}$ .

Proof.

The first part of the lemma follows directly from the fact that

f(\eta)=f(\eta^{2})=f(\eta)f(\eta).

To prove part (ii), assume that $\zeta\in\mathrm{orth}(\eta)$ . We then have

f(\zeta)f(\eta)=f(\zeta\eta)=f(0)=0.

Since $f$ is a bijection, the converse is also true, that is, if $\zeta$ is such that $f(\zeta)\in\mathrm{orth}(f(\eta))$ , then $\zeta\in\mathrm{orth}(\eta)$ . Therefore, $\zeta\in\mathrm{orth}(\eta)$ if and only if $f(\zeta)\in\mathrm{orth}(f(\eta))$ . This, together with the fact that $f(0)=0$ , implies that

\#\mathrm{orth}(f(\eta))=\#\mathrm{orth}(\eta).

Suppose now that $\bm{\varepsilon}\in\mathcal{E}_{n}$ . Since $f(0)=0$ and $f$ is bijective, we have $f(\bm{\varepsilon})\neq 0$ . This implies

2^{2^{n-1}-1}=\#\mathrm{orth}(\bm{\varepsilon})=\#\mathrm{orth}(f(\bm{\varepsilon}))=2^{v(f(\bm{\varepsilon}))}

and thus $v(f(\bm{\varepsilon}))=2^{n-1}-1$ . We deduce $f(\bm{\varepsilon})\in\mathcal{E}_{n}$ . ∎

Knowing how $f$ acts on $\mathbf{i_{1}}$ and on $\mathcal{E}_{n}$ , we can show the following.

Lemma 3.5.

Write $\bm{\varepsilon_{j}},\,1\leq j\leq 2^{n-1}$ , for the elements of $\mathcal{E}_{n}$ . Suppose that $f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ with

\displaystyle h=\sum_{j=1}^{2^{n-1}}\eta_{\bm{\varepsilon_{j}}}\bm{\varepsilon_{j}}.

Suppose that $f(\bm{\varepsilon_{j}})=\bm{\varepsilon_{k}}$ . Then

(3.2)

\displaystyle f(\mathbf{i_{1}}\bm{\varepsilon_{j}})=\mathbf{i_{1}}\eta_{\bm{\varepsilon_{k}}}\bm{\varepsilon_{k}}.

Furthermore, the function $f$ is completely determined by its action on the set $\mathbf{i_{1}}\mathcal{E}_{n}$ .

Proof.

The formula (3.2) follows from direct computation and from the orthogonality of the elements of the set $\mathcal{E}_{n}$ .

Suppose that we know the action of $f$ on the set $\mathbf{i_{1}}\mathcal{E}_{n}$ . Then, from the identity

\displaystyle f(\bm{\varepsilon_{k}})=-\big{(}f(\mathbf{i_{1}}\bm{\varepsilon_{k}})\big{)}^{2},

we can recover the value of $f(\bm{\varepsilon_{k}})$ . The identity

\displaystyle f(\mathbf{i_{1}})=\sum_{k=1}^{2^{n-1}}f(\mathbf{i_{1}}\bm{\varepsilon_{k}})

allows us to recover the value of $f(\mathbf{i_{1}})$ . The result then follows from Proposition 3.2. ∎

We now have all the tools to prove our Theorem 1.1.

3.2. Proof of Theorem 1.1 and its corollaries

Recall that we write $B_{n}$ for the set of signed permutations of length $n$ , which are permutations of $\{1,2,\ldots,n\}$ written in one-line notation where each entry may have a bar over it. An alternative description of $B_{n}$ , which is more useful to us, can be given as follows. Any signed permutation $\pi$ can be seen as a bijection of $\{1,\ldots,n,-1,\ldots,-n\}$ to itself such that $\pi(-i)=-\pi(i)$ and $-(-i)=i$ for $i=1,\ldots,n$ , where we have identified the bar with the $-$ sign. For a real number $a$ , we define $\mathrm{sgn}(a)=1$ if $a>0$ , $\mathrm{sng}(0)=0$ and $\mathrm{sgn}(a)=-1$ if $a<0$ .

Proof of Theorem 1.1.

Let $f$ be a real-linear automorphism of $\mathbb{M}\mathbb{C}(n)$ . From Lemma 3.5, $f$ is determined by its action on the set $\mathbf{i_{1}}\mathcal{E}_{n}$ and $f(\mathbf{i_{1}}\bm{\varepsilon_{j}})=\mathbf{i_{1}}\eta_{\bm{\varepsilon_{k}}}\bm{\varepsilon_{k}}$ , with $\eta_{\bm{\varepsilon_{k}}}\in\{-1,1\}$ . To such a function $f$ we can associate the signed permutation $\pi$ that satisfies $\pi(j)=\eta_{\bm{\varepsilon_{k}}}k$ . Conversely, for a given signed permutation $\pi$ , we can define the function $f$ by $f(\mathbf{i_{1}}\bm{\varepsilon_{j}})=\mathbf{i_{1}}\mathrm{sgn}(\pi(j))\bm{\varepsilon}_{|\pi(j)|}$ . It is clear from our construction that this bijection maps the identity to the identity and is compatible with composition. ∎

Theorem 1.1 allows us to deduce the corollaries that were presented in the Introduction.

Proof of Corollary 1.4.

For a given permutation of $\{1,\ldots,n\}$ , we can define a signed permutation by choosing whether we put a bar or not over each entry. It is thus obvious that $\#B_{n}=2^{n}\cdot n!$ . The result follows. ∎

Remark.

A more direct way to interpret the previous formula for the number of real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ corresponds to choosing a value for $f(\mathbf{i_{1}})$ and a permutation of $\mathcal{E}_{n}$ . There are $(2^{n-1})!$ such permutations, and since $f(\mathbf{i_{1}})^{2}=-1$ , from Proposition 1.2 there are $2^{2^{n-1}}$ possible values for $f(\mathbf{i_{1}})$ .

Proof of Corollary 1.5.

The formula in (i) and the asymptotic formula in (iii) follow from known results for signed involutions (see [5] and [15], respectively).

For (ii), by Theorem 1.1 it suffices to prove that $g(n)$ counts the number of signed involutions of length $n$ . But $\pi\in B_{n}$ either fixes $n$ and sends it to $n$ or $-n$ , or it sends $n$ to $j$ or $-j$ , where $j\in\{1,2,\ldots,n-1\}$ . Establishing the base cases $g(1)$ and $g(2)$ is straightforward. ∎

3.3. Counting involutions again

In this section, we present a more direct alternative approach to obtain the formula for the number of involutions of $\mathbb{M}\mathbb{C}(n)$ in Corollary 1.5 (i), using a counting argument. The main reason we choose to include this second proof is that this approach gives more insight into the nature of the multicomplex numbers since it relies on representation theorems for important subrings of $\mathbb{M}\mathbb{C}(n)$ .

Recall from Corollary 3.3 that $f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ for a unique $h\in H_{n}$ that depends on $f$ . Assuming that $f$ is an involution, the next result gives a way to choose $h$ .

Lemma 3.6.

Let $f$ be an involution of $\mathbb{M}\mathbb{C}(n)$ for $n\geq 1$ . Suppose that $f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ for some $h\in H_{n}$ . Then, we have $f(h)=h$ .

Proof.

Apply $f$ on $f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ to get

\displaystyle\mathbf{i_{1}}=f(\mathbf{i_{1}})f(h).

Using again $f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ , we see that

(3.3)

\displaystyle\mathbf{i_{1}}=\mathbf{i_{1}}hf(h).

Since $\mathbf{i_{1}}$ is invertible in $\mathbb{M}\mathbb{C}(n)$ , we obtain $1=hf(h)$ . Multiplying by $h$ , we therefore obtain $h=f(h)$ . ∎

Based on this last lemma, we introduce the sets

(3.4)

\displaystyle Y_{n}:=\left\{\sum_{h\in H_{n}}r_{h}h\,:\,r_{h}\in\mathbb{R}\right\}\quad\text{and}\quad\operatorname{fix}(f):=\{\eta\in Y_{n}\,:\,f(\eta)=\eta\}.

It is easy to see that $Y_{n}$ is a vector subspace and a subring of $\mathbb{M}\mathbb{C}(n)$ containing $H_{n}$ and $\operatorname{fix}(f)$ is a vector subspace and a subring of $Y_{n}$ if $f$ is an involution.

We can now prove the formula for $F(n)$ again.

Second proof of Corollary 1.5 (i).

Let $f$ be an involution of $\mathbb{M}\mathbb{C}(n)$ . We still have that $f$ is determined by its action on $\mathbf{i_{1}}$ and on $\mathcal{E}_{n}$ , and that it induces a permutation of the elements of this set. For $n=1$ , we already know that there are $2$ involutions on the complex space. This matches with the formula.

For $n\geq 2$ , since $f$ is an involution, the permutation induced by $f$ should contain cycles of length $2$ (transpositions) and should fix some elements of $\mathcal{E}_{n}$ . Let $k\in\{0,1,\ldots,2^{n-2}\}$ . The number of permutations of a set of $2^{n-1}$ elements with $k$ transpositions and $2^{n-1}-2k$ fixed elements is given by

\displaystyle\frac{(2^{n-1})!}{2^{k}k!(2^{n-1}-2k)!}.

We now need to find the possible values of $f(\mathbf{i_{1}})$ . We know that $f(\mathbf{i_{1}})=\mathbf{i_{1}}h$ for some choice of $h\in\operatorname{fix}(f)$ by Corollary 3.3 and Lemma 3.6. Since $\operatorname{fix}(f)$ is a subring of $Y_{n}$ and $f$ is real-linear, it is sufficient to know the coefficients of $h$ in its representation with respect to a basis of $\operatorname{fix}(f)$ . We will find such a basis. Denote the elements of $\mathcal{E}_{n}$ by $\bm{\varepsilon_{1}}$ , $\ldots$ , $\bm{\varepsilon_{2^{n-1}}}$ . Suppose that $f$ is given by

\displaystyle f(\bm{\varepsilon_{j_{1}}})=\bm{\varepsilon_{j_{2}}},\,f(\bm{\varepsilon_{j_{3}}})=\bm{\varepsilon_{j_{4}}},\,\ldots,\,f(\bm{\varepsilon_{j_{2k-1}}})=\bm{\varepsilon_{2k}}

and $f(\bm{\varepsilon_{j_{2k+1}}})=\bm{\varepsilon_{j_{2k+1}}}$ , $\ldots$ , $f(\bm{\varepsilon_{j_{2^{n-1}}}})=\bm{\varepsilon_{j_{2^{n-1}}}}$ . From Corollary 2.3 (ii), we can write

\displaystyle h=\sum_{\ell=1}^{2^{n-1}}c_{\ell}\bm{\varepsilon_{j_{\ell}}}

with $c_{\ell}\in\{-1,1\}$ . We have

\displaystyle f(h)=(c_{2}\bm{\varepsilon_{j_{1}}}+c_{1}\bm{\varepsilon_{j_{2}}})+\ldots+(c_{2k}\bm{\varepsilon_{j_{2k-1}}}+c_{2k-1}\bm{\varepsilon_{j_{2k}}})+c_{2k+1}\bm{\varepsilon_{j_{2k+1}}}+\ldots+c_{2^{n-1}}\bm{\varepsilon_{j_{2^{n-1}}}}.

It follows that $f(h)=h$ if and only if $c_{1}=c_{2},c_{3}=c_{4},\ldots,c_{2k-1}=c_{2k}$ . This is equivalent to stating that

\displaystyle\operatorname{fix}(f)=\mathrm{span}_{\mathbb{R}}\{(\bm{\varepsilon_{j_{1}}}+\bm{\varepsilon_{j_{2}}}),\,\ldots,\,(\bm{\varepsilon_{j_{2k-1}}}+\bm{\varepsilon_{j_{2k}}}),\bm{\varepsilon_{j_{2k+1}}},\,\ldots,\,\bm{\varepsilon_{j_{2^{n-1}}}}\}.

We deduce that $h$ should be of the form

\displaystyle h=\sum_{\underset{\ell\text{ odd}}{\ell=1}}^{2k-1}a_{\ell}(\bm{\varepsilon_{j_{\ell}}}+\bm{\varepsilon_{j_{\ell+1}}})+\sum_{\ell=2\ell+1}^{2^{n-1}}a_{\ell}\bm{\varepsilon_{j_{\ell}}},

with $a_{\ell}\in\{-1,1\}$ . This implies that the number of ways to choose $h$ is

\displaystyle 2^{2^{n-1}-k}.

Finally, summing from $k=0$ to $k=2^{n-2}$ , we obtain that the number of involutions of $\mathbb{M}\mathbb{C}(n)$ for $n\geq 2$ is

\displaystyle\sum_{k=0}^{2^{n-2}}\frac{2^{2^{n-1}-k}(2^{n-1})!}{2^{k}k!(2^{n-1}-2k)!}=(2^{n-1})!\sum_{k=0}^{2^{n-2}}\frac{2^{2^{n-1}-2k}}{k!(2^{n-1}-2k)!}.

This completes the proof. ∎

4. Solutions to $f^{(r)}=\mathrm{Id}$

In the last section, we investigated involutions, functions $f$ such that $f^{(2)}=\mathrm{Id}$ . The content of Corollary 1.6 is to answer the natural question of what happens when 2 is replaced by an arbitrary integer $r$ . Recall that we write $F_{r}(n)$ for the number of real-linear automorphisms of $\mathbb{M}\mathbb{C}(n)$ such that $f^{(r)}=\mathrm{Id}.$ From Theorem 1.1, the set of real-linear automorphisms $f$ of $\mathbb{M}\mathbb{C}(n)$ such that $f^{(r)}=\mathrm{Id}$ can be identified with the set of signed permutations $\pi\in B_{2^{n-1}}$ such that $\pi^{(r)}=\mathrm{Id}$ . We will thus enumerate such signed permutations in order to prove Corollary 1.6. Note that this proves directly Corollary 1.7 as well.

Proof of Corollary 1.6.

For a signed permutation $\pi$ , we let $\sigma=\mathrm{un}(\pi)$ stand for the corresponding unsigned permutation. Clearly, $\pi^{(r)}=\mathrm{Id}$ implies $\sigma^{(r)}=\mathrm{Id}$ while the converse is not true. Recall that $S_{n,t}$ stands for the set of permutations $\sigma$ of $n$ elements such that $\sigma^{(t)}=\mathrm{Id}$ . We have directly

\displaystyle F_{r}(n)=\sum_{\sigma\in S_{2^{n-1},r}}\#\left\{\pi\in B_{2^{n-1}}:\mathrm{un}(\pi)=\sigma,\pi^{(r)}=\mathrm{Id}\right\}.

The cardinalities of the sets in the above sum correspond to the number of ways of choosing the signs of the signed permutations. Any unsigned permutation $\sigma$ can be written as a product of disjoint cycles. Under the assumption that $\sigma^{(r)}=\mathrm{Id}$ , we have that the lengths of these cycles are divisors of $r$ . Assume that $\sigma$ has a cycle of length $s$ and assume without loss of generality that this cycle is $(1,2,\ldots,s)$ . Let $c_{1},c_{2},\ldots,c_{s}$ be the signs associated to the elements $1,2,\ldots,s$ in the signed permutation $\pi$ . We have

\displaystyle\pi^{(r)}(j)=(c_{1}\cdots c_{s})^{r/s}j,\qquad 1\leq j\leq s,

and thus

\displaystyle\pi^{(r)}(j)=j\text{ for }1\leq j\leq s\quad\iff\quad(c_{1}\cdots c_{s})^{r/s}=1.

If $r/s$ is even, then the signs $c_{1},\ldots,c_{s}$ can be chosen arbitrarily and the number of possible choices is equal to $2^{s}$ . On the other hand, if $r/s$ is odd then $c_{1}\cdots c_{s}$ must be equal to 1. The number of possible choices in this case is thus $2^{s-1}$ . For a given unsigned permutation $\sigma$ , we denote by $\mathrm{cyc}_{s}(\sigma)$ the number of disjoint cycles of length $s$ in $\sigma$ . We thus have

(4.1)

\#\left\{\pi\in B_{2^{n-1}}:\mathrm{un}(\pi)=\sigma,\pi^{(r)}=\mathrm{Id}\right\}=\left(\prod_{s|r,r/s\text{ is even }}2^{s\cdot\mathrm{cyc}_{s}(\sigma)}\right)\left(\prod_{s|r,r/s\text{ is odd }}2^{(s-1)\mathrm{cyc}_{s}(\sigma)}\right).

By noticing that for a fixed permutation $\sigma$ , $\sum_{s|r}s\cdot\mathrm{cyc}_{s}(\sigma)=2^{n-1}$ , the last expression can be rewritten as

(4.2)

\displaystyle\#\left\{\pi\in B_{2^{n-1}}:\mathrm{un}(\pi)=\sigma,\pi^{(r)}=\mathrm{Id}\right\}=2^{2^{n-1}}\left(\prod_{s|r,r/s\text{ is odd }}\frac{1}{2^{\mathrm{cyc}_{s}(\sigma)}}\right).

Summing over all permutations $\sigma\in S_{2^{n-1},r}$ , we get

\displaystyle F_{r}(n)=2^{2^{n-1}}\sum_{\sigma\in S_{2^{n-1},r}}\left(\prod_{s|r,r/s\text{ is odd }}\frac{1}{2^{\mathrm{cyc}_{s}(\sigma)}}\right).

Remark that when $r=p$ is an odd prime, equation (4.1) simplifies to

(4.3)

\#\left\{\pi\in B_{2^{n-1}}:\mathrm{un}(\pi)=\sigma,\pi^{(p)}=\mathrm{Id}\right\}=\Big{(}2^{(1-1)\mathrm{cyc}_{1}(\sigma)}\Big{)}\Big{(}2^{(p-1)\mathrm{cyc}_{p}(\sigma)}\Big{)}=2^{(p-1)\mathrm{cyc}_{p}(\sigma)}.

The number of permutations of $2^{n-1}$ elements with $k$ cycles of length $p$ and $2^{n-1}-pk$ fixed elements is given by

(4.4)

\frac{(2^{n-1})!}{p^{k}k!(2^{n-1}-pk)!}.

From equations (4.3) and (4.4), we conclude that for an odd prime $p$ ,

\displaystyle F_{p}(n)=(2^{n-1})!\sum_{k=0}^{\lfloor 2^{n-1}/p\rfloor}\frac{2^{k(p-1)}}{p^{k}k!(2^{n-1}-pk)!},

and this concludes the proof. ∎

Note that the above argument does not work when $r=2$ , although equation (4.4) still holds. In this case, equation (4.2) simplifies to

(4.5)

\displaystyle\#\left\{\pi\in B_{2^{n-1}}:\mathrm{un}(\pi)=\sigma,\pi^{(2)}=\mathrm{Id}\right\}=2^{2^{n-1}}\frac{1}{2^{\mathrm{cyc}_{2}(\sigma)}}=2^{2^{n-1}-k},

where $k$ is the number of cycles of length $2$ in the unsigned permutation $\sigma$ . From equations (4.4) and (4.5), we therefore obtain

F(n)=F_{2}(n)=(2^{n-1})!\sum_{k=0}^{2^{n-2}}\frac{2^{2^{n-1}-k}}{k!2^{k}(2^{n-1}-2k)!}=(2^{n-1})!\sum_{k=0}^{2^{n-2}}\frac{2^{2^{n-1}-2k}}{k!(2^{n-1}-2k)!},

as before.

5. Involutions preserving elementary units

Our focus is now on proving Theorem 1.8. The method of proof will be quite different from those used in previous sections. This comes from the fact that it is hard to devise a workable condition to detect if an $h\in H_{n}$ is an element of $\mathbb{I}(n)$ based on the idempotent components of $h$ . For this section, it will be more useful to use the canonical representation (2.2) of a multicomplex number.

If $f$ is an involution, then for any multicomplex number $\eta$ , we have

\displaystyle f(\eta)=\sum_{\mathbf{i}\in\mathbb{I}(n)}\eta_{\mathbf{i}}f(\mathbf{i}).

Since each $\mathbf{i}\in\mathbb{I}(n)\backslash\{1\}$ is a product of the units $\mathbf{i_{1}},\mathbf{i_{2}},\ldots,\mathbf{i_{n}}$ and $f$ is a real-linear homomorphism, we obtain the following proposition.

Proposition 5.1.

If $f:\mathbb{M}\mathbb{C}(n)\rightarrow\mathbb{M}\mathbb{C}(n)$ is an involution, then its values are completely determined by its action on $\mathbf{i_{1}},\mathbf{i_{2}},\ldots,\mathbf{i_{n}}$ .

Now, what are the possible values of each $f(\mathbf{i_{k}})$ for $1\leq k\leq n$ ? Since $f(\mathbf{i_{k}})\in U_{n}$ and here we restrict our attention to involutions preserving $\mathbb{I}(n)$ , $f(\mathbf{i_{k}})$ should be a product of an odd number of imaginary units $\mathbf{i_{1}}$ , $\mathbf{i_{2}}$ , $\ldots$ , $\mathbf{i_{n}}$ . Therefore, an $\mathbb{I}(n)$ -preserving involution $f$ of $\mathbb{M}\mathbb{C}(n)$ can be characterized by

f(\mathbf{i_{j}})=\mathbf{i_{1}}^{a_{1,j}}\mathbf{i_{2}}^{a_{2,j}}\cdots\mathbf{i_{n}}^{a_{n,j}}(-1)^{a_{n+1,j}},\qquad\,1\leq j\leq n.

Furthermore, for $1\leq j\leq n$ we have

-1=f(-1)=f(\mathbf{i_{j}^{2}})=f(\mathbf{i_{j}})^{2}=(-1)^{a_{1,j}}(-1)^{a_{2,j}}\cdots(-1)^{a_{n,j}},

which implies

\sum_{k=1}^{n}a_{k,j}\equiv 1\pmod{2},\qquad 1\leq j\leq n.

The first equality in the above chain of equalities comes from the fact that $f$ is, in particular, a ring homomorphism. In summary, we are trying to count functions $f$ on $\mathbb{M}\mathbb{C}(n)$ satisfying the following conditions:

(1)

$f(-1)=-1$ ;
(2)

$\mathbf{i_{j}^{2}}=-1$ for $1\leq j\leq n$ ;
(3)

$f(\mathbf{i_{j}}\mathbf{i_{k}})=f(\mathbf{i_{j}})f(\mathbf{i_{k}})$ for $1\leq j,k\leq n$ ;
(4)

$f(f(\mathbf{i_{j}}))=\mathbf{i_{j}}$ for $1\leq j\leq n$ ;
(5)

$f(\mathbf{i_{j}})=\mathbf{i_{1}}^{a_{1,j}}\mathbf{i_{2}}^{a_{2,j}}\cdots\mathbf{i_{n}}^{a_{n,j}}(-1)^{a_{n+1,j}}$ for $1\leq j\leq n$ ;
(6)

$a_{1,j},\ldots,a_{n+1,j}\in\{0,1\}$ and $\sum_{k=1}^{n}a_{k,j}\equiv 1\pmod{2}$ for $1\leq j\leq n$ .

With this setup, we are now ready to prove Theorem 1.8.

Proof of Theorem 1.8.

We will use a matrix representation to obtain the number of involutions satisfying the previous description. A function $f$ as described by $(1)$ to $(6)$ above will be represented by the matrix

A_{f}:=\begin{bmatrix}a_{1,1}&a_{1,2}&\ldots&a_{1,n}&0\\ a_{2,1}&a_{2,2}&\ldots&a_{2,n}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_{n,1}&a_{n,2}&\cdots&a_{n,n}&0\\ a_{n+1,1}&a_{n+1,2}&\cdots&a_{n+1,n}&1\end{bmatrix}.

The condition $\sum_{k=1}^{n}a_{k,j}\equiv 1\pmod{2}$ for $1\leq j\leq n$ then translates to a condition on the matrix $A_{f}$ as

(5.1)

\begin{bmatrix}a_{1,1}&a_{1,2}&\ldots&a_{1,n}&0\\ a_{2,1}&a_{2,2}&\ldots&a_{2,n}&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_{n,1}&a_{n,2}&\ldots&a_{n,n}&0\\ a_{n+1,1}&a_{n+1,2}&\ldots&a_{n+1,n}&1\end{bmatrix}^{T}\begin{bmatrix}1\\ 1\\ 1\\ 1\\ 0\end{bmatrix}\equiv\begin{bmatrix}1\\ 1\\ 1\\ 1\\ 0\end{bmatrix}\pmod{2}.

To simplify the notation, we will simply write $A$ for $A_{f}$ . On the other hand, the condition $f(f(\mathbf{i_{j}}))=\mathbf{i_{j}},\,1\leq j\leq n$ , translates into

(5.2)

A^{2}\equiv I\pmod{2}.

Our problem thus becomes the problem of enumerating matrices $A$ with $\{0,1\}$ entries and satisfying conditions (5.1) and (5.2).

We set $X=A-I$ . Since we are working modulo 2, we have

A^{2}\equiv I\Leftrightarrow A^{2}-I\equiv 0\pmod{2}\Leftrightarrow(A-I)^{2}\equiv 0\pmod{2}\Leftrightarrow X^{2}\equiv 0\pmod{2}.

The problem now becomes enumerating $(n+1)\times(n+1)$ matrices $X$ such that

(1)

The entries of $X$ are equal to 0 or 1,
(2)

$X^{2}\equiv 0\pmod{2}$ ;
(3)

The sum of each column of $X$ is $\equiv 0\pmod{2}$ ;
(4)

The right column of $X$ has only zeros.

We denote by $Y$ the submatrix of $X$ obtained by omitting the right column of $X$ and its bottom row, that is

Y=\begin{bmatrix}a_{1,1}-1&a_{1,2}&\ldots&a_{1,n}\\ a_{2,1}&a_{2,2}-1&\ldots&a_{2,n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n,1}&a_{n,2}&\ldots&a_{n,n}-1\end{bmatrix}.

The conditions on $X$ imply the following conditions on $Y$ :

(1)

The entries of $Y$ are equal to 0 or 1;
(2)

$Y^{2}\equiv 0\pmod{2}$ ;
(3)

The sum of each column of $Y$ is $\equiv 0\pmod{2}$ .

We denote by $k$ the dimension of the kernel of $Y$ , that is, $k:=\dim(\ker(Y))$ . Because $Y^{2}=0$ , we have

(5.3)

k\geq n/2.

Observe that the dimension of the kernel of $Y$ is equal to the dimension of the kernel of $Y^{\top}$ . It will be easier to work with this transpose.

We use the notation $\vec{e}:=(1,1,\ldots,1)^{\top}$ . Condition 3 on the matrix $Y$ is equivalent to

\vec{e}\in\ker(Y^{\top}).

For a fixed value of $k$ , the number of ways of choosing $\ker(Y^{\top})$ with the restriction that $\vec{e}\in\ker(Y^{\top})$ is given by

(5.4)

B(k,n):=\prod_{j=1}^{k-1}\frac{2^{n}-2^{j}}{2^{k}-2^{j}},

with the convention that $B(k,n)=1$ for $k=0,1$ . To see this, first note that the number of ways of choosing an ordered sequence of $k$ linearly independent vectors (with $\vec{e}$ as the first vector of the sequence) is given by

(5.5)

\prod_{j=1}^{k-1}(2^{n}-2^{j}),

since when choosing a new vector, one cannot choose any linear combination of previously chosen vectors. Now, many choices of vector sequences (or basis choices) will describe the same subspace. Given a basis of linearly independent vectors, the number of ways of choosing a basis that will span the same subspace (under the condition that $\vec{e}$ is the first vector of the ordered basis) is given by

(5.6)

\prod_{j=1}^{k-1}(2^{k}-2^{j}).

Equality (5.4) follows from (5.5) and (5.6).

Now, suppose that the kernel $\ker(Y^{\top})$ has been chosen. Let $k$ again be the dimension of the kernel of $Y^{\top}$ . Let $\vec{u}_{1},\vec{u}_{2},\ldots\vec{u}_{k}$ be a vector basis of $\ker(Y^{\top})$ . Let $\vec{v}_{1},\vec{v}_{2},\ldots,\vec{v}_{n-k}$ be vectors such that $\vec{u}_{1},\vec{u}_{2},\ldots\vec{u}_{k},\vec{v}_{1},\vec{v}_{2},\ldots,\vec{v}_{n-k}$ is a vector basis of $\mathbb{Z}_{2}^{n}$ . We will now find the number of ways of choosing $\mathrm{Ran}\,(Y^{\top})$ , the range of $Y^{\top}$ .

Since $Y^{2}\equiv 0\pmod{2}$ , we deduce that $(Y^{\top})^{2}\equiv 0\pmod{2}$ . This last identity implies that

(Y^{\top})^{2}\vec{v}_{j}=\vec{0},\qquad 1\leq j\leq n-k

and thus

Y^{\top}\vec{v}_{j}\in\ker(Y^{\top}),\qquad 1\leq j\leq n-k.

We therefore have

Y^{\top}\vec{v}_{j}=\sum_{s=1}^{k}r_{s,j}\vec{u}_{s},

where $r_{s,j}\in\{0,1\}$ and $1\leq j\leq n-k$ . The number of ways to choose the value of $Y^{\top}\vec{v}_{1}$ is given by

2^{k}-1.

This comes from the fact that $\vec{v}_{1}\not\in\ker(Y^{\top})$ and therefore the $r_{s,j}$ cannot all be zero. One can now choose the values of $Y^{\top}\vec{v}_{2},Y^{\top}\vec{v}_{3},\ldots,Y^{\top}\vec{v}_{n-k}$ under the restriction that the vectors $Y^{\top}\vec{v}_{j}$ must be linearly independent. To see why these vectors must be linearly independent, suppose that $\vec{w}$ is a linear combination of the vectors $\vec{v}_{j}$ such that $Y^{\top}\vec{w}=\vec{0}$ . This implies $\vec{w}\in\ker(Y^{\top})$ . We thus have exhibited a vector $\vec{w}$ that can be expressed both as a linear combination of the vectors $\vec{u}_{j}$ and as a linear combination of the vectors $\vec{v}_{j}$ . This contradicts the fact that $\vec{u}_{1},\vec{u}_{2},\ldots\vec{u}_{k},\vec{v}_{1},\vec{v}_{2},\ldots,\vec{v}_{n-k}$ is a basis of $\mathbb{Z}_{2}^{n}$ . This constraint of linear independence implies that the number of ways of choosing the vector $\vec{v}_{j}$ is equal to $2^{k}-2^{j-1}$ for $1\leq j\leq n-k$ . We conclude that the number of ways of choosing the image of $Y^{\top}$ is given by

(5.7)

D(k,n)=\prod_{j=0}^{n-k-1}(2^{k}-2^{j}).

Putting everything together, the number of ways of choosing the matrix $Y^{\top}$ , or equivalently the number of ways of choosing the matrix $Y$ , is given by

B(k,n)D(k,n)=\prod_{j=1}^{k-1}\frac{2^{n}-2^{j}}{2^{k}-2^{j}}\prod_{j=0}^{n-k-1}(2^{k}-2^{j}).

Finally, to fully specify the matrix $X$ , one has to specify its bottom row (describing the signs in the involution). The condition $X^{2}\equiv 0\pmod{2}$ can be written, after transposing, as

\begin{bmatrix}a_{1,1}-1&a_{2,1}&\cdots&a_{n,1}&a_{n+1,1}\\ a_{1,2}&a_{2,2}-1&\cdots&a_{n,2}&a_{n+1,2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_{1,n}&a_{2,n}&\cdots&a_{n,n}-1&a_{n+1,n}\\ 0&0&\cdots&0&0\end{bmatrix}\begin{bmatrix}a_{1,1}-1&a_{2,1}&\cdots&a_{n,1}&a_{n+1,1}\\ a_{1,2}&a_{2,2}-1&\cdots&a_{n,2}&a_{n+1,2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_{1,n}&a_{2,n}&\cdots&a_{n,n}-1&a_{n+1,n}\\ 0&0&\cdots&0&0\end{bmatrix}\equiv 0

modulo 2. This implies, in particular, that

$\displaystyle(a_{1,1}-1,a_{2,1},\ldots,a_{n,1})\cdot(a_{n+1,1},a_{n+1,2},\ldots,a_{n+1,n})$	$\displaystyle\equiv$	$\displaystyle 0\pmod{2},$
$\displaystyle(a_{1,2},a_{2,2}-1,\ldots,a_{n,2})\cdot(a_{n+1,1},a_{n+1,2},\ldots,a_{n+1,n})$	$\displaystyle\equiv$	$\displaystyle 0\pmod{2},$
	$\displaystyle\vdots$
$\displaystyle(a_{1,n},a_{2,n},\ldots,a_{n,n}-1)\cdot(a_{n+1,1},a_{n+1,2},\ldots,a_{n+1,n})$	$\displaystyle\equiv$	$\displaystyle 0\pmod{2}.$

It follows that the vector

\vec{w}:=\begin{bmatrix}a_{n+1,1}\\ a_{n+1,2}\\ \vdots\\ a_{n+1,n}\end{bmatrix}

must be in the kernel of the matrix

\begin{bmatrix}a_{1,1}-1&a_{2,1}&\ldots&a_{n,1}\\ a_{1,2}&a_{2,2}-1&\ldots&a_{n,2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1,n}&a_{2,n}&\ldots&a_{n,n}-1\end{bmatrix},

which is precisely $Y^{\top}$ . If the dimension of the kernel of $Y^{\top}$ is equal to $k$ , then the number of ways of choosing the sign column

\begin{bmatrix}a_{n+1,1}\\ a_{n+1,2}\\ \vdots\\ a_{n+1,n}\end{bmatrix}

is equal to

(5.8)

2^{k}.

This last number comes from the fact that $\vec{w}=\sum_{s=1}^{k}r_{s}\vec{u}_{k}$ with $r_{s}\in\{0,1\}$ .

From equations (5.3), (5.4), (5.7), and (5.8), we conclude that the number of involutions satisfying (1) to (6) is equal to

\sum_{n/2\leq k\leq n}D(k,n)B(k,n)2^{k}=\sum_{n/2\leq k\leq n}\Big{(}\prod_{j=1}^{k-1}\frac{2^{n}-2^{j}}{2^{k}-2^{j}}\Big{)}\Big{(}\prod_{j=0}^{n-k-1}2^{k}-2^{j}\Big{)}2^{k}.

This completes the proof. ∎

6. Generating $r$ -Involutions

The proof of Corollary 1.6 combined with the explicit bijection in the proof of Theorem 1.1 gives a way to generate the $r$ -involutions of $\mathbb{M}\mathbb{C}(n)$ , for $r\geq 2$ . Here, we describe this method (which is a brute force method).

(1)

Select a permutation $\sigma\in S_{2^{n-1},r}$ of the symbols $\{1,2,\ldots,2^{n-1}\}$ .
(2)

Generate all the possible sign permutations $\pi$ by considering all the possible sign insertions in $\sigma$ .
(3)

For a given sign permutation in the last step, check if $\pi^{(r)}=\mathrm{Id}$ .
(4)

Generate the $r$ -involution of $\mathbb{M}\mathbb{C}(n)$ by setting $f(\mathbf{i_{1}}\bm{\varepsilon_{j}})=\mathbf{i_{1}}\mathrm{sgn}(\pi(j))\bm{\varepsilon_{|\pi(j)|}}$ .

Using this method we can generate, for example, the following $6$ -involution of the space $\mathbb{M}\mathbb{C}(3)$ . Letting $\bm{\varepsilon_{1}}=(1-\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}})/4$ , $\bm{\varepsilon_{2}}=(1+\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}-\mathbf{i_{2}}\mathbf{i_{3}})/4$ , $\bm{\varepsilon_{3}}=(1-\mathbf{i_{1}}\mathbf{i_{2}}+\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}})/4$ , and $\bm{\varepsilon_{4}}=(1+\mathbf{i_{1}}\mathbf{i_{2}}-\mathbf{i_{1}}\mathbf{i_{3}}+\mathbf{i_{2}}\mathbf{i_{3}})/4$ , then

f(\mathbf{i_{1}}\bm{\varepsilon_{1}})=\mathbf{i_{1}}\bm{\varepsilon_{3}},\,f(\mathbf{i_{1}}\bm{\varepsilon_{2}})=\mathbf{i_{1}}(-1)\bm{\varepsilon_{2}},\,f(\mathbf{i_{1}}\bm{\varepsilon_{3}})=\mathbf{i_{1}}\bm{\varepsilon_{4}},\,f(\mathbf{i_{1}}\bm{\varepsilon_{4}})=\mathbf{i_{1}}\bm{\varepsilon_{1}}.

This $6$ -involution comes from the following sign permutation (using the bar notation):

\displaystyle\pi=3\overline{2}41.

Note that we could create a $3$ -involution by using the unsigned permutation

\displaystyle\sigma=3241.

We can rewrite the above $6$ -involution $f$ using the elementary units (canonical representation) as follows:

	$\displaystyle f(\eta)=$	$\displaystyle\,\eta_{1}+\mathbf{i_{1}}(\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}+\eta_{\mathbf{i_{1}}}+\eta_{\mathbf{i_{2}}}+\eta_{\mathbf{i_{3}}})/2+\mathbf{i_{2}}(-\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}+\eta_{\mathbf{i_{1}}}-\eta_{\mathbf{i_{2}}}+\eta_{\mathbf{i_{3}}})/2$
		$\displaystyle+\mathbf{i_{1}}\mathbf{i_{2}}\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}+\mathbf{i_{3}}(\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}+\eta_{\mathbf{i_{1}}}-\eta_{\mathbf{i_{2}}}-\eta_{\mathbf{i_{3}}})/2-\mathbf{i_{1}}\mathbf{i_{3}}\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}-\mathbf{i_{2}}\mathbf{i_{3}}\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}$
		$\displaystyle+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}(-\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}+\eta_{\mathbf{i_{1}}}+\eta_{\mathbf{i_{2}}}-\eta_{\mathbf{i_{3}}})/2,$

where

\displaystyle\eta=\eta_{1}+\mathbf{i_{1}}\eta_{\mathbf{i_{1}}}+\mathbf{i_{2}}\eta_{\mathbf{i_{2}}}+\mathbf{i_{1}}\mathbf{i_{2}}\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}+\mathbf{i_{3}}\eta_{\mathbf{i_{3}}}+\mathbf{i_{1}}\mathbf{i_{3}}\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}+\mathbf{i_{2}}\mathbf{i_{3}}\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}.

This $6$ -involution is non-trivial in the sense that it sends some of the elementary units in $\mathbb{I}(3)$ to units in $U_{3}\backslash\mathbb{I}(3)$ . From the expression above, we see that

f(\mathbf{i_{1}})=\frac{\mathbf{i_{1}}+\mathbf{i_{2}}+\mathbf{i_{3}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2},\;f(\mathbf{i_{2}})=\frac{\mathbf{i_{1}}-\mathbf{i_{2}}-\mathbf{i_{3}}+\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2},\text{ and }f(\mathbf{i_{3}})=\frac{\mathbf{i_{1}}+\mathbf{i_{2}}-\mathbf{i_{3}}-\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}{2}.

The proof of Theorem 1.8 suggests a way to generate a list of $\mathbb{I}(n)$ -preserving involutions for a fixed value of $n$ .

(1)

Fix a value of $k$ and loop over the values of $k\in[n/2,n]$ .
(2)

Generate a basis $\vec{u}_{1},\ldots\vec{u}_{k}$ (including $\vec{e}$ ) of all subspaces of dimension $k$ of $\mathbb{Z}_{2}^{n}$ .
(3)

For each basis in the list generated in step 2, find a set of vectors $\vec{v}_{1},\ldots,\vec{v}_{n-k}$ so that the vectors $\vec{u}_{1},\ldots,\vec{u}_{k},\vec{v}_{1},\ldots,\vec{v}_{n-k}$ form a basis of $\mathbb{Z}_{2}^{n}$ .
(4)

For each vector $\vec{v}_{j}$ , choose the image of $\vec{v}_{j}$ as a linear combination of the vectors $\vec{u}_{j}$ . Note this image by $\vec{s}_{j}$ .

(5)

Obtain the matrix $Y^{\top}$ as in the above proof by solving

Y^{\top}[\vec{u}_{1},\ldots,\vec{u}_{k},\vec{v}_{1},\ldots,\vec{v}_{n-k}]=[\vec{0},\ldots,\vec{0},\vec{s}_{1},\ldots,\vec{s}_{n-k}].

(6)

Generate a list of involutions by looping over all possibilities for the choice of the sign vector associated to $Y$ , fully specifying the matrix $X$ in the above proof.

Here is a sample of such involutions preserving elementary units for $n=3$ :

(1)

$f(\eta)=\eta_{1}-\eta_{\mathbf{i_{1}}}\mathbf{i_{1}}-\eta_{\mathbf{i_{2}}}\mathbf{i_{2}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}\mathbf{i_{1}i_{2}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{3}}+\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{1}i_{3}}+\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}\mathbf{i_{2}i_{3}}+\eta_{\mathbf{i_{3}}}\mathbf{i_{1}i_{2}i_{3}}$ ;
(2)

$f(\eta)=\eta_{1}-\eta_{\mathbf{i_{1}}}\mathbf{i_{1}}-\eta_{\mathbf{i_{2}}}\mathbf{i_{2}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}\mathbf{i_{1}i_{2}}-\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{3}}-\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{1}i_{3}}-\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}\mathbf{i_{2}i_{3}}-\eta_{\mathbf{i_{3}}}\mathbf{i_{1}i_{2}i_{3}}$ ;
(3)

$f(\eta)=\eta_{1}+\eta_{\mathbf{i_{1}}}\mathbf{i_{1}}+\eta_{\mathbf{i_{3}}}\mathbf{i_{2}}+\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}\mathbf{i_{1}i_{2}}+\eta_{\mathbf{i_{2}}}\mathbf{i_{3}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}}\mathbf{i_{1}i_{3}}+\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{2}i_{3}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{1}i_{2}i_{3}}$ ;
(4)

$f(\eta)=\eta_{1}+\eta_{\mathbf{i_{1}}}\mathbf{i_{1}}-\eta_{\mathbf{i_{3}}}\mathbf{i_{2}}-\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}\mathbf{i_{1}i_{2}}-\eta_{\mathbf{i_{2}}}\mathbf{i_{3}}-\eta_{\mathbf{i_{1}}\mathbf{i_{3}}}\mathbf{i_{1}i_{3}}+\eta_{\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{2}i_{3}}+\eta_{\mathbf{i_{1}}\mathbf{i_{2}}\mathbf{i_{3}}}\mathbf{i_{1}i_{2}i_{3}}$ .

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Counting involutions on multicomplex spaces

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

Theorem 1.1.

Proposition 1.2.

Proposition 1.3.

Corollary 1.4.

Corollary 1.5.

Remark.

Corollary 1.6.

Corollary 1.7.

Theorem 1.8.

2. Background on multicomplex numbers

2.1. Multicomplex numbers

2.2. An idempotent representation for multicomplex numbers

2.3. Representation theorems and bijections

Proposition 2.1.

Proof.

Proof of Proposition 1.2.

Proposition 2.2.

Proof.

Corollary 2.3.

Proof.

3. Characterization of Involutions of 𝕄​ℂ​(n)\mathbb{M}\mathbb{C}(n)

Definition 3.1.

3.1. Auxiliary results

Proposition 3.2.

Corollary 3.3.

Proof.

Lemma 3.4.

Proof.

Lemma 3.5.

Proof.

3.2. Proof of Theorem 1.1 and its corollaries

Proof of Theorem 1.1.

Proof of Corollary 1.4.

Remark.

Proof of Corollary 1.5.

3.3. Counting involutions again

Lemma 3.6.

Proof.

Second proof of Corollary 1.5(i).

4. Solutions to f(r)=Idf^{(r)}=\mathrm{Id}

Proof of Corollary 1.6.

5. Involutions preserving elementary units

Proposition 5.1.

Proof of Theorem 1.8.

6. Generating rr-Involutions

References

3. Characterization of Involutions of $\mathbb{M}\mathbb{C}(n)$

Second proof of Corollary 1.5 (i).

4. Solutions to $f^{(r)}=\mathrm{Id}$

6. Generating $r$ -Involutions